To solve the inequality \(|2x + 1| < |x|\), consider the cases for the absolute values.

Case 1: When \(x \geq 0\), \(|x| = x\). The inequality becomes \(|2x + 1| < x\).

Subcase 1.1: If \(2x + 1 \geq 0\), then \(|2x + 1| = 2x + 1\). The inequality becomes \(2x + 1 < x\), which simplifies to \(x < -1\). This is not possible since \(x \geq 0\).

Subcase 1.2: If \(2x + 1 < 0\), then \(|2x + 1| = -(2x + 1)\). The inequality becomes \(-(2x + 1) < x\), which simplifies to \(-2x - 1 < x\) or \(-1 < 3x\), giving \(x > -\frac{1}{3}\). However, this contradicts \(x \geq 0\).

Case 2: When \(x < 0\), \(|x| = -x\). The inequality becomes \(|2x + 1| < -x\).

Subcase 2.1: If \(2x + 1 \geq 0\), then \(|2x + 1| = 2x + 1\). The inequality becomes \(2x + 1 < -x\), which simplifies to \(3x < -1\), giving \(x < -\frac{1}{3}\).

Subcase 2.2: If \(2x + 1 < 0\), then \(|2x + 1| = -(2x + 1)\). The inequality becomes \(-(2x + 1) < -x\), which simplifies to \(-2x - 1 < -x\) or \(-1 < x\), giving \(x > -1\).

Combining the results from Case 2, we have \(-1 < x < -\frac{1}{3}\).